Question

find the tangent plane to the surface of revolution obtained by rotating y = e^x about the x axis at the point (0, sqrt(2)/2,sqrt(2)/2)

Answer 1

so ive got the parametrization of the surface as \[<x, e^xcos(\theta),e^xsin(\theta)>\]

Answer 2

With the above i get x = 0 and theta = pi/4

Answer 3

i get two vectors as \[<0,\cos(\theta),\sin(\theta)>\] and \[<x, \frac{\sqrt{2}}{2}e^x, \frac{\sqrt{2}}{2}e^x>\]

Answer 4

derivative of first vector \[<0,-\sin(\theta),\cos(\theta)>\] and second one \[<1, \frac{\sqrt{2}}{2}e^x, \frac{\sqrt{2}}{2}e^x>\]

Answer 5

evaluate first vector at pi/4\[<0, - \frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2}>\] and second one a x = 0\[<1, \frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2}>\]

Answer 6

take cross product and get \[-i - \frac{\sqrt{2}}{2}j + \frac{\sqrt{2}}{2}k\]

Answer 7

Equation of the plane is \[-x - \frac{\sqrt{2}}{2}y + \frac{\sqrt{2}}{2}z = 0\]

Answer 8

Is that correct?

Answer 9

@SolomonZelman @texaschic101 ?

Answer 10

what do you think @dumbcow ?

Answer 11

@Directrix ?

Answer 12

ok sorry i wanted to make sure i was doing it right.

you did cross-product correctly but something else is off, why did you use vector
<0, cos, sin> ?

the tangent plane is defined using the gradient (derivative)
\[\delta_x (x - x_0) + \delta_y(y - y_0) + \delta_z(z - z_0) = 0\]
for point (x0,y0,z0)
derivative gives :
\[<1, \frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2}>\]
thus equation is
\[x + \frac{\sqrt{2}}{2}(y-\frac{\sqrt{2}}{2}) + \frac{\sqrt{2}}{2} (z - \frac{\sqrt{2}}{2}) = 0\]

Answer 13

i got the two vectors by letting one variable be constant and letting the other variable vary so with that vector i let x be constant and theta vary which gives that vector.

Answer 14

and the opposite for the other vector, that way i can cross product the two and get a normal vector.

Answer 15

the locus
\[y^2 + z^2 = e ^{2x}\]
\[ø = y^2 + z^2 - e ^{2x}\]
\[grad (ø) = < -2 e ^{2x}, 2y, 2z > = n ^{->}\]
\[n ^{->} (0, \sqrt{2}/2, \sqrt{2}/2) = <-2, \sqrt{2}, \sqrt{2} > \]
\[π: - 2x + \sqrt{2}y + \sqrt{2} z = K\]

\[K = 0 + \sqrt{2} • \sqrt{2}/2 + \sqrt{2} • \sqrt{2}/2 = 2\]

\[π: -\sqrt 2x + y + z = \sqrt 2\]